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A coil 3.75 cm radius, containing 580 turns, is placed in a uniform magnetic field that varies...

Question:

A coil 3.75 cm radius, containing 580 turns, is placed in a uniform magnetic field that varies with time according to

{eq}\ B= (1.20 *10^{-2} T/s) t+(3.15*10^{-5} T/s^4 )t^4 {/eq}. The coil is connected to a 590-{eq}\Omega {/eq} resistor, and its plane is perpendicular to the magnetic field. You can ignore the resistance of the coil.

Part A : Find the magnitude of the induced emf in the coil as a function of time.

{eq}- E= 9.79*10^{-3} V +( 1.03*10^{-4} V/s^3) t^3 \\ - E= 3.07*10^{-2} V +( 8.07*10^{-5} V/s^3) t^3 \\ - E= 3.07*10^{-2} V +( 3.23*10^{-4} V/s^3) t^3 \\ - E= 9.79*10^{-3} V +( 3.23*10^{-4} V/s^3) t^3 {/eq}

Part B : What is the current in the resistor at time t0 = 4.70 s ?

Faraday's Law of Electromagnetic Induction:

{eq}\\ {/eq}

Electromagnetic induction is the phenomenon in which a changing magnetic field in a region induces an electric field in that region. Thus, if a closed-loop is placed in a region of a changing magnetic field, then an e.m.f is induced in the loop.

The magnitude of the induced emf depends on the number of turns in the loop and how the magnetic flux through the loop is changing with time. The dependence is described quantitatively by Faraday's Law of Electromagnetic Induction.

Answer and Explanation:

{eq}\\ {/eq}

We are given:

  • The radius of coil, {eq}r=3.75\;\rm cm=3.75\times 10^{-2}\;\rm m {/eq}.
  • The number of turns in the coil, {eq}N=580 {/eq}.
  • The magnetic field as a function of time, {eq}B= (1.20 \times 10^{-2} \;{\rm T/s}) t+(3.15\times 10^{-5} \;{\rm T/s^4 })t^4 {/eq}.
  • The angle between the plane of the loop and the magnetic field, {eq}\phi=90^{\circ} {/eq}.
  • The resistance connected to the coil, {eq}R=590\;\rm \Omega {/eq}.


PART (A):

The angle between the normal to the loop and the magnetic field is:

{eq}\begin{align*} \theta&=90^{\circ}-\phi\\ &=90^{\circ}-90^{\circ}\\ &=0^{\circ} \end{align*} {/eq}


According to Faraday's Law of Electromagnetic Induction, if the flux of the magnetic flux associated with a closed-loop is changing with time, then the magnitude of e.m.f induced in the loop is given by the following equation:

{eq}\varepsilon=N\dfrac{d \phi}{dt} {/eq}

Here,

  • {eq}\phi {/eq} is the magnetic flux through the closed-loop.
  • {eq}\varepsilon {/eq} is the emf induced in the loop.
  • {eq}N {/eq} is the number of turns in the loop.


The magnetic flux through a loop of the area, {eq}A {/eq} with a magnetic field, {eq}B {/eq}, is given by the equation:

{eq}\phi=BA\cos\,\theta {/eq}, where,

  • {eq}\theta {/eq} is the angle between the normal to the loop and the magnetic field.


Therefore, the e.m.f induced in the loop is:

{eq}\begin{align*} \varepsilon&=N\dfrac{d \phi}{dt}\\ &=N\dfrac{d \left (\pi r^2 B\cos\,\theta \right )}{dt}\\ &=N\pi r^2\cos\,\theta\dfrac{dB}{dt}\\ &=580\times \pi\times \left (3.75\times 10^{-2}\;\rm m \right )^2\times \cos\,0^{\circ}\times \left (\dfrac{d\left ((1.20 \times 10^{-2} \;{\rm T/s}) t+(3.15\times 10^{-5} \;{\rm T/s^4 })t^4 \right )}{dt} \right )\\ &=2.56\;{\rm m^2}\times \left ((1.20 \times 10^{-2} \;{\rm T/s})+4(3.15\times 10^{-5} \;{\rm T/s^4 })t^3 \right )\\ &=3.07\times 10^{-2}\;{\rm T.m^2/s}+\left (3.23\times 10^{-4}{\rm T.m^2/s^4 } \right )t^3\\ &=\boxed{3.07\times 10^{-2}\;{\rm V}+\left (3.23\times 10^{-4}{\rm V/s^3 } \right )t^3} \end{align*} {/eq}


PART (B):

At {eq}t=4.70\;\rm s {/eq}, the e.m.f in the circuit is:

{eq}\begin{align*} \varepsilon&=3.07\times 10^{-2}\;{\rm V}+\left (3.23\times 10^{-4}{\rm V/s^3 } \right )\left (4.70\;\rm s \right )^3\\ &=3.07\times 10^{-2}\;{\rm V}+\left (3.35\times 10^{-2}{\rm V } \right )\\ &=6.42\times 10^{-2}\;\rm V \end{align*} {/eq}


According to Ohm's Law, the current, {eq}I {/eq}, through a conductor depends on the resistance, {eq}R {/eq}, of the conductor, and the potential difference, {eq}V {/eq}, applied across the conductor by the following equation:

{eq}V=IR {/eq}

After plugging the given values into the above equation, we have:

{eq}\begin{align*} 6.42\times 10^{-2}\;{\rm V}&=I\times 590\;{\rm \Omega}\\ \Rightarrow I&=\dfrac{6.42\times 10^{-2}\;{\rm V}}{590\;{\rm \Omega}}\\ &=\boxed{1.09\times 10^{-4}\;\rm A} \end{align*} {/eq}



Learn more about this topic:

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Faraday's Law of Electromagnetic Induction: Equation and Application

from High School Physics: Help and Review

Chapter 13 / Lesson 10
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